Given that

A vertical pole fixed to the horizontal ground is divided in the ratio $3: 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 \mathrm{m}$ away from the base of the pole.

$\tan \theta =\frac{x}{6}$
$\tan 2\theta =\frac{10x}{18}=\frac{5x}{9}$

We know that $\tan 2\theta = \frac{2 \tan \theta }{1 - {\tan }^2\theta }$

$\frac{5x}{9}=\frac{\frac{x}{3}}{1-\frac{x^2}{36}}$

$\Rightarrow 5=\frac{3}{1-\frac{x^2}{36}}$

$1-\frac{x^2}{36}=\frac{3}{5}$

$\frac{x^2}{36}=\frac{2}{5}$

$\Rightarrow x^2=\frac{72}{5} \Rightarrow x=\frac{6\sqrt{2}}{\sqrt{5}}$

From the figure ; 

 Length of pole $=10\mathrm{x}=\frac{60\sqrt{2}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}$

$=\frac{60\sqrt{10}}{5}=12\sqrt{10}m$